Let $f$ be a continuous, real valued function defined on a closed, bounded interval $I$, a subset of real numbers. Let $\{f_n\}$ be a Cauchy sequence in the $L^2$ norm. Give a counterexample that the sequence does not converge in the $L^2$ norm metric.
Thank you so much for your help. You would save my life!!!!!!!!!!
As @RobertIsrael pointed out, a Cauchy sequence must converge to an $L^2$ function. However, if you are looking for a sequence that does not converge to a continuous function, think of something like this :
$\int |f|^2$ computes the area under the curve of $|f|^2$. Try and construct a sequence of functions $f_n \geq 0$ such that $\int f_n^2$ is the area of a triangle with base $1/n$, and height $1$ (Think of a sequence that "bumps up" at $1/2 \in [0,1]$).
Now try to show that this sequence is Cauchy in $L^2$ by explicitly computing $\int |f_n-f_m|^2$.
Show that it coverges point-wise to a function which is not continuous.